mark scheme june 2007 6663 core mathematics c1

7/21/2019 Mark Scheme June 2007 6663 Core Mathematics C1

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Mark Scheme (Final)

Summer 2007

GCE

GCE Mathematics (6663/01)

Edexcel Limited. Registered in England and Wales No. 4496750Registered Office: One90 High Holborn, London WC1V 7BH


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General Principal for Pure Mathematics Marking

(But note that specific mark schemes may sometimes override these general principles).

Method mark for solving 3 term quadratic:

1. Factorisationcpqqxpxcbxx =++=++ where),)(()( 2 , leading to x=

amncpqqnxpmxcbxax ==++=++ andwhere),)(()( 2 , leading to x=

2. FormulaAttempt to use correct formula (with values for a, band c).

3. Completing the square

Solving 02 =++ cbxx : 0,0,)( 2 qpcqpx , leading tox=

Method marks for differentiation and integration:1. Differentiation

Power of at least one term decreased by 1. ( 1 nn xx )

2. Integration

Power of at least one term increased by 1. ( 1+ nn xx )

Use of a formulaWhere a method involves using a formula that has been learnt, the advice given in recent

examiners reports is that the formula should be quoted first.Normal marking procedure is as follows:Method mark for quoting a correct formula and attempting to use it, even if there are mistakes inthe substitution of values.Where the formula is not quoted, the method mark can be gained by implication from correctworking with values, but may be lost if there is any mistake in the working.

Exact answersExaminers reports have emphasised that where, for example, an exact answer is asked for, orworking with surds is clearly required, marks will normally be lost if the candidate resorts to usingrounded decimals.

Answers without workingThe rubric says that these may not gain full credit. Individual mark schemes will give details ofwhat happens in particular cases. General policy is that if it could be done in your head, detailedworking would not be required. Most candidates do show working, but there are occasionalawkward cases and if the mark scheme does not cover this, please contact your team leader foradvice.


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June 20076663 Core Mathematics C1

Mark Scheme

Question Scheme Marks

number

1. 9 5 or 23 3 5 3 5 5 5+ or23 5 5 or ( )223 5 M1

= 4 A1cso (2)

2

M1 for an attempt to multiply out. There must be at least 3 correct terms. Allow one sign slip

only, no arithmetic errors.

e.g. ( )223 3 5 3 5 5+ + is M1A0

( )2

23 3 5 3 5 5+ + is M1A0 as indeed is 9 6 5 5

BUT 9 15 15 5( 4)+ = is M0A0 since there is more than a sign error.

553536 + is M0A0 since there is an arithmetic error.

If all you see is 9 + 5 that is M1 but please check it has not come from incorrect working.

Expansion of ( )( )3 5 3 5+ + is M0A0

A1cso for 4 only. Please check that no incorrect working is seen.

Correct answer only scores both marks.


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number

2. (a) Attempt 3 8 or 3 4

)8( M1

= 16 A1 (2)

(b) 5

13

x 5,

13

x B1, B1 (2)4

(a) M1 for: 2 (on its own) or ( )4

3 32 or ( )4

43 38 or 8 or 2 or 3 4 38 or 4096

38 or 512 or ( )3

1

4096 is M0

A1 for 16 only

(b) 1stB1 for 5 on its own or something.

So e.g.x

x 34

5is B1 But

135 is B0

An expression showing cancelling is not sufficient

(see first expression of QC0184500123945 the mark is scored for the second expression)

2ndB1 for13x

Can use ISW (incorrect subsequent working)

e.g435x scores B1B0 but it may lead to

3 45x which we ignore as ISW.

Correct answers only score full marks in both parts.



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number

3. (a) 21

2

46

d

d 1 +=

xx

x

y or

126 2x x

+

M1 A1 (2)

(b) 23

6

+ x or3

26 1 x

+ M1 A1ft (2)

(c)

3

3 28

3x x C+ + A1: 3

3

3x or

2

3

4 23

x A1: both, simplified and + C M1 A1 A1 (3)

7

(a) M1 for some attempt to differentiate: 1n nx x

Condone missingd

d

y

xory=

A1 for both terms correct, as written or better. No + Chere. Of course2

xis acceptable.

(b) M1 for some attempt to differentiate again. Follow through theird

d

y

x, at least one term correct

or correct follow through.

A1f.t. as written or better, follow through must have 2 distinct terms and simplified e.g.4

14

= .

(c) M1 for some attempt to integrate: 1n nx x + . Condone misreading2

2

d dor

d d

y y

x xfory.

(+Calone is not sufficient)

1stA1 for either 33

3x or

2

3

4 23

x(or better)

32

24

3x is OK here too but not for 2ndA1.

2ndA1 for both 3x and32

8 8or

3 3

x x x i.e. simplified terms and +Call on one line.

322 instead of

3

8is OK


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number

4. (a) Identify a = 5 and d= 2 (May be implied) B1

( )200 (200 1)u a d= + (= 5 + (200 1)2 ) M1

= 403(p) or () 4.03 A1 (3)

(b) ( ) [ ] ( )200200 200

2 (200 1) or "their 403"2 2

S a d a= + + M1

[ ] ( )200 200

2 5 (200 1) 2 or 5 "their 403"2 2

= + + A1

= 40 800 or 408 A1 (3)6

(a) B1 can be implied if the correct answer is obtained. If 403 is not obtained then the values of

aand dmust be clearly identified as a= 5 and d= 2.

This mark can be awarded at any point.

M1 for attempt to use nth term formula with n= 200. Follow through their aand d.

Must have use of n= 200 and one of aor d correct or correct follow through.

Must be 199 not 200.

A1 for 403 or 4.03 (i.e. condone missing sign here). Condone 403 here.

N.B. a= 3, d= 2 is B0 and a+ 200dis M0 BUT 3 200 2+ is B1M1 and A1 if it leads to 403.

Answer only of 403 (or 4.03) scores 3/3.

(b) M1 for use of correct sum formula with n= 200. Follow through their aand dand their 403.

Must have some use of n= 200,and some of a, d or lcorrect or correct follow through.

1stA1 for any correct expression (i.e. must have a= 5 and d= 2) but can f.t. their 403 still.

2ndA1 for 40800 or 408 (i.e. the sign is required before we accept 408 this time).

40800p is fine for A1 but 40800 is A0.

ALT Listing

(a) They might score B1 if a=5 and d= 2 are clearly identified. Then award M1A1 together for 403.

(b) =

+200

1

)32(r

r . Give M1 for k3)201(2

2002 + (with k>1), A1 for k= 200 and A1 for 40800.


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Question Scheme Marksnumber

5. (a)

Translation parallel tox-axis M1

Top branch intersects +vey-axisLower branch has no intersections A1

No obvious overlap

2

3,0 or

3

2marked ony- axis B1 (3)

(b) 0,2 == yx B1, B1 (2)

S.C. [Allow ft on first B1 forx= 2 when translated the wrong way but must becompatible with their sketch.]

5

(a) M1 for a horizontal translation two branches with one branch cuttingy axis only.

If one of the branches cuts both axes (translation up and across) this is M0.

A1 for a horizontal translation to left. Ignore any figures on axes for this mark.

B1 for correct intersection on positivey-axis. More than 1 intersection is B0.

x=0 andy= 1.5 in a table alone is insufficient unless intersection of their sketch is with +vey-axis.

A point marked on the graph overrides a point given elsewhere.

(b) 1stB1 for 2x= . NBx 2 is B0.

Can acceptx= +2 if this is compatible with their sketch.

Usually they will have M1A0 in part (a) (and usually B0 too)

2ndB1 fory= 0.

S.C. Ifx= -2 andy=0 and some other asymptotes are also given award B1B0

The asymptote equations should be clearly stated in part (b). Simply markingx=-2 ory=0

on the sketch is insufficient unless they are clearly marked asymptotex= -2 etc.

x

y


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number

6. (a) 8)4(2 2 = xxx M1

0842 =+ xx (*) A1cso (2)

(b) ( )2

81444 2 =x or ( )2

2 4 8 0x + = M1

x= - 2 + (any correct expression) A1

3431648 == or 12 4 3 2 3= = B1

) 4322 =y M: Attempt at least oneyvalue M1

326,322 +=+= yx 326,322 == yx A1 (5)

7

(a) M1 for correct attempt to form an equation inxonly. Condone sign errors/slips but attempt at

this line must be seen. E.g. 2 22 4 8x x x = is OK for M1.

A1cso for correctly simplifying to printed form. No incorrect working seen. The = 0 is required.

These two marks can be scored in part (b). For multiple attempts pick best.

(b) 1stM1 for use of correct formula. If formula is not quoted then a fully correct substitution is

required. Condone missingx= or just + or instead of + for M1.

For completing the square must have as printed or better.

If they have 2 4 8 0x x = then M1 can be given for ( )2

2 4 8 0x = .

1stA1 for -2 + any correct expression. (The + is required butx= is not)

B1 for simplifying the surd e.g. 48 4 3= . Must reduce to b 3 so 16 3 or 4 3 are OK.

2ndM1 for attempting to find at least oneyvalue. Substitution into one of the given equations

and an attempt to solve fory.

2ndA1 for correctyanswers. Pairings need not be explicit but they must say which isxand whichy.

Mis-labellingxandyloses final A1 only.



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number

7. (a) Attempt to use discriminant acb 42 M1

01240)3(4 22 >>+ kkkk (*) A1cso (2)

(b) 01242 = kk

( )( ) ( ) ( )2

2 24 4 4 12, with 12 or or 2 2 122

k a k b ab k k = = M1

k =2 and 6 (both) A1

6,2 >< kk or ( ) ( ), 2 ; 6 , M: choosing outside M1 A1ft (4)

6

(a) M1 for use of acb 42 , one of bor cmust be correct.

Or full attempt using completing the square that leads to a 3TQ in k

e.g.

2 2

( 3)2 4k kx k

+ = +

A1cso Correct argument to printed result. Need to state (or imply) that acb 42 >0 and no

incorrect working seen. Must have >0. If > 0 just appears with ( )2 4 3 0k k + > that is OK.If >0 appears on last line only with no explanation give A0.

acb 42 followed by 01242 > kk only is insufficient so M0A0

e.g. 2 4 1 3k k + (missing brackets) can get M1A0 but 2 4( 3)k k+ + is M0A0 (wrong formula)

Using 2 4 0b ac > is M0.

(b) 1stM1 for attempting to find critical regions. Factors, formula or completing the square.

1stA1 for k= 6 and 2 only2ndM1 for choosing the outside regions

2ndA1f.t. as printed or f.t. their (non identical) critical values

6 < k< 2 is M1A0 but ignore if it follows a correct version2 < k< 6 is M0A0 whatever their diagram looks like

Condone use ofxinstead of kfor critical values and final answers in (b).

Treat this question as 3 two mark parts. If part (a) is seen in (b) or vice versa marks can be awarded.


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8. (a) 53)( 2 += ka [must be seen in part (a) or labelled 2a = ] B1 (1)

(b) 5)53(3)( 3 ++= ka M1

= 9k+ 20 (*) A1cso (2)

(c)(i) )6527(5)209(34 +=++= kka M1

)6527()209()53(

4

1

++++++==

kkkka

r

r M1

(ii) = 40k+ 90 A1

= 10(4k+ 9) (or explain why divisible by 10) A1ft (4)

7

(b) M1 for attempting to find 3a , follow through their 2a k .

A1cso for simplifying to printed result with no incorrect working seen.

(c) 1stM1 for attempting to find 4a . Can allow a slip here e.g. 3(9k+ 20) [i.e. forgot +5]

2ndM1 for attempting sum of 4 relevant terms, follow through their (a) and (b).

Must have 4 terms starting with k.

Use of arithmetic series formulae at this point is M0A0A0

1stA1 for simplifying to 40k+ 90 or better

2ndA1ft for taking out a factor of 10 or dividing by 10 or an explanation in words true k .

Follow through their sum of 4 terms provided that both Ms are

scored and their sum is divisible by 10.

A comment is not required.

e.g.40 90

4 910

kk

+= + is OK for this final A1.

S.C. =

5

2r

ra = 120k+ 290 = 10(12k+ 29) can have M1M0A0A1ft.


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number

9. (a) )(122

10

3

6)(f

23

Cxxx

x += M1 A1

x= 5: 6560125250 =+ C C= 0 M1 A1 (4)

(b) )1252( 2 xxx or )4)(32( 2 + xxx or )4)(32( 2 xxx + M1

= )4)(32( + xxx (*) A1cso (2)

(c)

Shape B1

Through origin B1

)0,4(and0,2

3

B1 (3)

9

(a) 1stM1 for attempting to integrate, 1n nx x +

1stA1 for allxterms correct, need not be simplified. Ignore + Chere.

2ndM1 for some use ofx= 5 and f(5)=65 to form an equation in Cbased on their integration.

There must be some visible attempt to usex= 5 and f(5)=65. No +Cis M0.

2ndA1 for C= 0. This mark cannot be scored unless a suitable equation is seen.

(b) M1 for attempting to take out a correct factor or to verify. Allow usual errors on signs.

They must get to the equivalent of one of the given partially factorised expressions or, if

verifying, 2(2 3 8 12)x x x x+ i.e. with no errors in signs.

A1cso for proceeding to printed answer with no incorrect working seen. Comment not required.

This mark is dependent upon a fully correct solution to part (a) so M1A1M0A0M1A0 for (a) & (b).

Will be common or M1A1M1A0M1A0. To score 2 in (b) they must score 4 in (a).

(c) 1stB1 for positive 3x shaped curve (with a max and a min) positioned anywhere.

2ndB1 for any curve that passes through the origin (B0 if it only touches at the origin)

3rdB1 for the two points clearly given as coords or values marked in appropriate places onxaxis.

Ignore any extra crossing points (they should have lost first B1).

Condone (1.5, 0) if clearly marked on vex-axis. Condone (0, 4) etc if marked on +vexaxis.

Curve can stop (i.e. not pass through) at (-1.5, 0) and (4, 0).

A point on the graph overrides coordinates given elsewhere.

x

y


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number

10. (a) 145:1 =+== yx , 14216:2 =+== yx (can be given 1stB1 for 1

in (b) or (c)) 2nd B1 for - 14

( ) ( ) 170)1(1412 22 =+=PQ (*) M1 A1cso (4)

(b) 123 46 += xxxy M1

22 4123d

d = xxxx

y M1 A1

134123d

d:1 ===

x

yx M: Evaluate at one of the points M1

1312412d

d:2 ===

x

yx Parallel A: Both correct + conclusion A1 (5)

(c) Finding gradient of normal

=

131m M1

)1(13

11 = xy M1 A1ft

01413 = yx o.e. A1cso (4)

13

(a) M1 for attempting PQ or 2PQ using their Pand their Q. Usual rules about quoting formulae.

We must see attempt at ( )22

1 QP yy + for M1.2

...PQ = etc could be M1A0.A1cso for proceeding to the correct answer with no incorrect working seen.

(b) 1stM1 for multiplying by 2x , the 3 2 or 6x x must be correct.2ndM1 for some correct differentiation, at least one term must be correct as printed.

1stA1 for a fully correct derivative.

These 3 marks can be awarded anywhere when first seen.

3rdM1 for attempting to substitutex= 1 orx= 2 in their derivative. Substituting inyis M0.2ndA1 for -13 from both substitutions and a brief comment.

The 13 must come from their derivative.

(c) 1

st

M1 for use of the perpendicular gradient rule. Follow through their 13.2ndM1 for full method to find the equation of the normal or tangent at P. If formula is

quoted allow slips in substitution, otherwise a correct substitution is required.

1stA1ft for a correct expression. Follow through their 1 and their changed gradient.

2ndA1cso for a correct equation with = 0 and integer coefficients.

This mark is dependent upon the 13 coming from their derivative in (b) hence cso.

Tangent can get M0M1A0A0, changed gradient can get M0M1A1A0orM1M1A1A0.

Condone confusion over terminology of tangent and normal, mark gradient and equation.

MR Allow forx

4 or (x+6) but not omitting 14 x or treating it as 4x.


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11. (a)3

( 4)2

y x= + Gradient =2

3 M1 A1 (2)

(b) 42323 +=+ xx x=, 9

4 M1, A1

==+

=

3

13

3

102

9

43y A1 (3)

(c) Wherey= 1, 2:3

1: 21 == BA xlxl M: Attempt one of these M1 A1

Area = )1)((2

1 PAB yxx M1

1 7 7 49 132

2 3 3 18 18= = = o.e. A1 (4)

9

(a) M1 for an attempt to write 3x+ 2y 8 =0 in the formy= mx+ c

or a full method that leads to m= , e.g find 2 points, and attempt gradient using 2 1

2 1

y y

x x

e.g. findingy= -1.5x+ 4 alone can score M1 (even if they go on to say m= 4)

A1 for m=3

2

(can ignore the +c) or2

3

d

d=

x

y

(b) M1 for forming a suitable equation in one variable and attempting to solve leading tox= ..ory=

1stA1 for any exact correct value forx

2ndA1 for any exact correct value fory

(These 3 marks can be scored anywhere, they may treat (a) and (b) as a single part)

(c) 1stM1 for attempting thexcoordinate ofAorB. One correct value seen scores M1.

1stA1 for 13 and 2A Bx x= =

2ndM1 for a full method for the area of the triangle follow through their , ,A B Px x y .

e.g. determinant approach1 43 91 1 1

2 2 3103

2 2 2 ... ( ...)1 1 1

=

2ndA1 for49

18or an exact equivalent.

All accuracy marks require answers as single fractions or mixed numbers not necessarily in lowest

terms.

mark scheme june 2007 6663 core mathematics c1

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